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STRUCTURE OF Cu-63 AND Cu-65
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF Cu-63 WITH S = -3/2 For getting the structure of Cu-63 we use the structure of Ni -64. (See my STRUCTURE OF Ni-60 AND Ni-64 ). In the following diagram you see the two alpha particles on the left side of Mg-24 with the deuterons p21n21, n22p22, p23n23, and n24p24. On the right side of the Mg-24 you see also the two alpha particle with the deuterons n25p25, p26n26, n27p27 and p28n28. However the deuterons p13n13, n14p14, p15n15, and n16p16 existing in front of Mg-24 (from the second horizontal plane to the fifth horizontal plane) are not shown. Also the deuterons n17p17, p18n18, n19p19, and p20n20 are not shown because they are behind the Mg-24. Note that all these deuterons from p1n1 to p28n28 give spin S=0 . For getting the structure of Cu-63 we add the additional deuteron p29n29, which fills the blank positions formed by the deuterons n11p12 and n24p24. It has spin S = -1. Here the extra n30(-1/2) contributes for getting a struture of symmetry, becuase it fills the symmetrical position near p11. Whereas in the nuclear structure of Cu-58 the additional p28n28 With S =+1 breaks the symmetry. Then in the structure of Cu-63 adding the two extra neutrons n31(+1/2) and n32(+1/2) one gets S = 0. Thus the total spin S = -3/2 of the Cu-63 is due to the summation of spins of the three extra neutrons like the n30(-1/2), the n33(-1/2) and the n34(-1/2). Of course they form a stable structure because they make also np bonds along the spin axis with strong binding energies. Note that the n33 and n34 are not shown here because the n33 is behind the p12 , while the n34 is in front of the p11 '''. '''DIAGRAM OF Cu-63 WITH S = -3/2 ' ' ' n29………p12..........n12' ' p29……… n11.........p11….n30 Sixth horizontal plane' ' p24..........n10........p10…… n28' ' n24………..p9...........n9 ……p28 Fifth horizontal plane' ' n23.........p8..........n8............p27' ' p23..........n7...........p7...........n27 Fourth horizontal plane ' ' p22.........n6...........p6............n26' ' n22……….p5...........n5………p26 Third horizontal plane ' ' n21………p4..........n4………….p25' ' p21……..n3………p3………..n25 Second horizontal plane' ' n2………..p2…………n32' ' n31………p1.........n1 First horizontal plane' ' ' STRUCTURE OF THE STABLE Cu-65 WITH S = -3/2 According to the experiments Copper (Cu) has two stable isotopes, Cu-63 and Cu-65, along with 27 radioisotopes. Since Cu-65 has the same spin S =-3/2 as that of Cu-63 we get the structure of Cu-65 and the spin S = -3/2 by adding two additional deuterons of opposite spin like the n35(+1/2) and the n36(-1/2). Note that the n35 is in front of p1(+1/2) and makes the two np bonds like(n35-p1) and the (n35-p13). Such a bond is very strong because n35 is coupled with p13 along the spin axis . The p13 is not shown in the above diagram because it is in front of n3. On the other hand the n36 fills the blank position between the p27 and p19 which is behind the n8. These extra np bonds contribute more to the increase of binding energy able to overcome the repulsive energy of pp systems with repulsive forces of long range. Category:Fundamental physics concepts